Saved Bookmarks
| 1. |
An electrical device draws 2 kW power form AC mains [voltage 223 V (rms) = sqrt(50,000) V]. The current differs (lags) in phase by phi (tan = (-3)/(4)) as compared to voltage. Find (i) R, (ii) X_(C ) - X_(L), and (iii) I_(M). Another device has twice the values for R, X_(C ) and X_(L). How are the answers affected ? |
|
Answer» Solution :`P = 2KW = 2000W, tan phi = (-3)/(4) , I_(m)=I_(0) ? R = ? X_(C)- X(L)=?` `V_("rms") = V= 223 V` `Z = (V^2)/(P) = 25 OMEGA` `Z = sqrt(R^2 + (X_(L) - X_(C ))^2)` `625 = R^2 + (X_(L) - X+_(C ))^2` Again `tan phi = (X_L - X_C)/ (R ) = (3)/(4)` `X_(L) - X_( C) = (3R)/(4)` using this `R= 20 Omega, X_(L) - X_( C) = 15 Omega, I = (V)/ (Z) = (223)/(25) = 8.92 A` `I_(m) = sqrt(2) I = 12.6 A` |
|